Linear transformation with polynomial












1












$begingroup$


Determine if its linear the transformation



$f:Re_{n}[x] rightarrow Re$



such as



$f(p(x))=p(x)+1$



for any $p(x) in Re_{n}$[x]



The solution says it's not linear but I worked it out like this:



Condition 1. $T(u+v)=T(u)+T(v)$



$f(p(x))+p'(x))=((p(x)+1)+(p'(x)+1))=f(p(x))+f(p'(x))$



Condition 2. $alpha T(u)=T(alpha u)$



$alpha f(p(x))= alpha (p(x)+1) = alpha p(x)+alpha = f(alpha p(x))$



What am I doing wrong?






linear-transformations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Take a very careful look at the first equality of condition 1
    $endgroup$
    – user328442
    Jan 5 at 21:44












  • $begingroup$
    Your fourth line contradicts your second one: the $;f;$ you're defining there is not a function to $;Bbb R;$ (as written in second line) as it doesn't give a number (=a scalar), but a new polynomial. It also confusing what you wrote in the line below "condition 1", with derivatives and etc. Where does this all come from? So what exactly is the counter domain of $;f;$ and what exactly is the definition of $;f;$ ?
    $endgroup$
    – DonAntonio
    Jan 5 at 22:11










  • $begingroup$
    My notation was unfortunate, the p' is not the derivative
    $endgroup$
    – Jakcjones
    Jan 5 at 22:40












  • $begingroup$
    As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
    $endgroup$
    – Jakcjones
    Jan 5 at 22:42


















1












$begingroup$


Determine if its linear the transformation



$f:Re_{n}[x] rightarrow Re$



such as



$f(p(x))=p(x)+1$



for any $p(x) in Re_{n}$[x]



The solution says it's not linear but I worked it out like this:



Condition 1. $T(u+v)=T(u)+T(v)$



$f(p(x))+p'(x))=((p(x)+1)+(p'(x)+1))=f(p(x))+f(p'(x))$



Condition 2. $alpha T(u)=T(alpha u)$



$alpha f(p(x))= alpha (p(x)+1) = alpha p(x)+alpha = f(alpha p(x))$



What am I doing wrong?






linear-transformations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Take a very careful look at the first equality of condition 1
    $endgroup$
    – user328442
    Jan 5 at 21:44












  • $begingroup$
    Your fourth line contradicts your second one: the $;f;$ you're defining there is not a function to $;Bbb R;$ (as written in second line) as it doesn't give a number (=a scalar), but a new polynomial. It also confusing what you wrote in the line below "condition 1", with derivatives and etc. Where does this all come from? So what exactly is the counter domain of $;f;$ and what exactly is the definition of $;f;$ ?
    $endgroup$
    – DonAntonio
    Jan 5 at 22:11










  • $begingroup$
    My notation was unfortunate, the p' is not the derivative
    $endgroup$
    – Jakcjones
    Jan 5 at 22:40












  • $begingroup$
    As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
    $endgroup$
    – Jakcjones
    Jan 5 at 22:42
















1












1








1





$begingroup$


Determine if its linear the transformation



$f:Re_{n}[x] rightarrow Re$



such as



$f(p(x))=p(x)+1$



for any $p(x) in Re_{n}$[x]



The solution says it's not linear but I worked it out like this:



Condition 1. $T(u+v)=T(u)+T(v)$



$f(p(x))+p'(x))=((p(x)+1)+(p'(x)+1))=f(p(x))+f(p'(x))$



Condition 2. $alpha T(u)=T(alpha u)$



$alpha f(p(x))= alpha (p(x)+1) = alpha p(x)+alpha = f(alpha p(x))$



What am I doing wrong?






linear-transformations







share|cite|improve this question











$endgroup$




Determine if its linear the transformation



$f:Re_{n}[x] rightarrow Re$



such as



$f(p(x))=p(x)+1$



for any $p(x) in Re_{n}$[x]



The solution says it's not linear but I worked it out like this:



Condition 1. $T(u+v)=T(u)+T(v)$



$f(p(x))+p'(x))=((p(x)+1)+(p'(x)+1))=f(p(x))+f(p'(x))$



Condition 2. $alpha T(u)=T(alpha u)$



$alpha f(p(x))= alpha (p(x)+1) = alpha p(x)+alpha = f(alpha p(x))$



What am I doing wrong?






linear-transformations




linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 22:37







Jakcjones

















asked Jan 5 at 21:40









JakcjonesJakcjones

828




828








  • 1




    $begingroup$
    Take a very careful look at the first equality of condition 1
    $endgroup$
    – user328442
    Jan 5 at 21:44












  • $begingroup$
    Your fourth line contradicts your second one: the $;f;$ you're defining there is not a function to $;Bbb R;$ (as written in second line) as it doesn't give a number (=a scalar), but a new polynomial. It also confusing what you wrote in the line below "condition 1", with derivatives and etc. Where does this all come from? So what exactly is the counter domain of $;f;$ and what exactly is the definition of $;f;$ ?
    $endgroup$
    – DonAntonio
    Jan 5 at 22:11










  • $begingroup$
    My notation was unfortunate, the p' is not the derivative
    $endgroup$
    – Jakcjones
    Jan 5 at 22:40












  • $begingroup$
    As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
    $endgroup$
    – Jakcjones
    Jan 5 at 22:42
















  • 1




    $begingroup$
    Take a very careful look at the first equality of condition 1
    $endgroup$
    – user328442
    Jan 5 at 21:44












  • $begingroup$
    Your fourth line contradicts your second one: the $;f;$ you're defining there is not a function to $;Bbb R;$ (as written in second line) as it doesn't give a number (=a scalar), but a new polynomial. It also confusing what you wrote in the line below "condition 1", with derivatives and etc. Where does this all come from? So what exactly is the counter domain of $;f;$ and what exactly is the definition of $;f;$ ?
    $endgroup$
    – DonAntonio
    Jan 5 at 22:11










  • $begingroup$
    My notation was unfortunate, the p' is not the derivative
    $endgroup$
    – Jakcjones
    Jan 5 at 22:40












  • $begingroup$
    As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
    $endgroup$
    – Jakcjones
    Jan 5 at 22:42










1




1




$begingroup$
Take a very careful look at the first equality of condition 1
$endgroup$
– user328442
Jan 5 at 21:44






$begingroup$
Take a very careful look at the first equality of condition 1
$endgroup$
– user328442
Jan 5 at 21:44














$begingroup$
Your fourth line contradicts your second one: the $;f;$ you're defining there is not a function to $;Bbb R;$ (as written in second line) as it doesn't give a number (=a scalar), but a new polynomial. It also confusing what you wrote in the line below "condition 1", with derivatives and etc. Where does this all come from? So what exactly is the counter domain of $;f;$ and what exactly is the definition of $;f;$ ?
$endgroup$
– DonAntonio
Jan 5 at 22:11




$begingroup$
Your fourth line contradicts your second one: the $;f;$ you're defining there is not a function to $;Bbb R;$ (as written in second line) as it doesn't give a number (=a scalar), but a new polynomial. It also confusing what you wrote in the line below "condition 1", with derivatives and etc. Where does this all come from? So what exactly is the counter domain of $;f;$ and what exactly is the definition of $;f;$ ?
$endgroup$
– DonAntonio
Jan 5 at 22:11












$begingroup$
My notation was unfortunate, the p' is not the derivative
$endgroup$
– Jakcjones
Jan 5 at 22:40






$begingroup$
My notation was unfortunate, the p' is not the derivative
$endgroup$
– Jakcjones
Jan 5 at 22:40














$begingroup$
As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
$endgroup$
– Jakcjones
Jan 5 at 22:42






$begingroup$
As for the $f: Re_{n} [x] rightarrow Re$, I see what your saying. Maybe is a mistake in the manual.
$endgroup$
– Jakcjones
Jan 5 at 22:42












1 Answer
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$begingroup$

It should be
$$f(p(x)+p'(x)) = p(x)+p'(x)+1 ne p(x)+p'(x)+2 = f(p(x)) + f(p'(x))$$



Anyway, it is faster to note that $f(0) = 1 ne 0$, so $f$ cannot be linear.






share|cite|improve this answer









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    $begingroup$

    It should be
    $$f(p(x)+p'(x)) = p(x)+p'(x)+1 ne p(x)+p'(x)+2 = f(p(x)) + f(p'(x))$$



    Anyway, it is faster to note that $f(0) = 1 ne 0$, so $f$ cannot be linear.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      It should be
      $$f(p(x)+p'(x)) = p(x)+p'(x)+1 ne p(x)+p'(x)+2 = f(p(x)) + f(p'(x))$$



      Anyway, it is faster to note that $f(0) = 1 ne 0$, so $f$ cannot be linear.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        It should be
        $$f(p(x)+p'(x)) = p(x)+p'(x)+1 ne p(x)+p'(x)+2 = f(p(x)) + f(p'(x))$$



        Anyway, it is faster to note that $f(0) = 1 ne 0$, so $f$ cannot be linear.






        share|cite|improve this answer









        $endgroup$



        It should be
        $$f(p(x)+p'(x)) = p(x)+p'(x)+1 ne p(x)+p'(x)+2 = f(p(x)) + f(p'(x))$$



        Anyway, it is faster to note that $f(0) = 1 ne 0$, so $f$ cannot be linear.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 21:44









        mechanodroidmechanodroid

        28.4k62548




        28.4k62548






























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